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Given any three matrices $\gamma_i \in M_2(\mathbb{C})$, $i=1,2,3$, which satisfy $$\gamma_i \gamma_j + \gamma_j \gamma_i = 2I \delta_{ij}$$ And given a vector $\mathbf{v} \in \mathbb{R}^3$, show that $$exp(i\mathbf{v} \cdot \gamma) = \cos(||\mathbf{v}||)I+ I\frac{\sin(||\mathbf{v}||)}{||\mathbf{v}||}\mathbf{v} \cdot \gamma$$

Where $\mathbf{v} \cdot \gamma = \Sigma_{i=1}^{i=3}v_i \gamma_i$. So I've shown that $$(\mathbf{v} \cdot \gamma)^2 = ||\mathbf{v}||^2 I$$ while trying to use the Taylor expansion to prove the required relation, which didn't go anywhere. The Taylor expansion I used was $exp(tA) = I + tA + \frac{t^2}{2}A^2...$ and the $t$ is there because it's usually helpful in group theory, which is where this question came up.

Unfortunately the internet suggests that what I have shown is wrong and it should be $(\mathbf{v} \cdot \gamma)^2 = ||\mathbf{v}||^2$. So I don't know where to go from here, any help/hints would be much appreciated!

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    This looks like a pretty straightforward generalization of Rodrigues' Formula; you might want to look up a proof of that and work from there. – 2017-02-03
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    Use that $e^Z = \sum_{k=0}\frac{Z^k}{k!} = \sum_{n=0}^\infty \frac{Z^{2n}}{(2n)!} + \frac{Z^{2n+1}}{(2n+1)!}$ and note that $(v\cdot \gamma)^{2n} = |v|^{2n}$ and $(v\cdot \gamma)^{2n+1} = |v|^{2n}(v\cdot\gamma)$ and relate this to the Taylor series for $\sin$ and $\cos$ and you'll get the desired formula. – 2017-02-04
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    Also note that $I$ is $1$ for all practical purposes. The identity-matrix acts just like $1$ does for real numbers. To be technically correct you should have $I$ there to denote that the expression is a matrix, however this is often just ignored. – 2017-02-04

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